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# Strength of Materials MCQs Bars of Composite Sections – 1

If a bar of sections of two different length and different diameters are in a line and P load is acting axially on them then what will be the change in length of the bar?

a) P/E x (L1 + L2)

b) P/E x (A1/L1 + A2/ L2)

c) P/E x (L1/A1 + L2/A2)

d) E/P x (L1/A1 + L2/A2)

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Answer: c

Explanation: Change in length of section 1 = PL1/EA1

Change in length of section 2 = PL2/EA2

Total change in length of bar = PL1/EA1 + PL2/EA2.

How does the elastic constant varys with the elongation of body?

a) The elastic constant is directly proportional to the elongation

b) The elastic constant is directly proportional to the elongation

c) The elongation does not depends on the elastic constant

d) None of these

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Answer: b

Explanation: Elongation of a composite bar of two sections = P/E x (L1/A1 + L2/A2)

E is inversely proportional to bar elongation.

A composite rod is 1000mm long, its two ends are 40 mm2 and 30mm2 in area and length are 400mm and 600mm respectively. If the rod is subjected to an axial tensile load of 1000N, what will be its total elongation(E = 200GPa)?

a) 0.130m

b) 0.197mm

c) 0.160mm

d) 0.150mm

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Answer: a

Explanation: As elongation of a composite bar of two sections = P/E x (L1/A1 + L2/A2)

Putting L1, L2, A1 and A2 400mm2, 600mm2, 40mm2 and 30mm2 and P = 1000 and E = 200 x 103.

A mild steel wire 5mm in diameter and 1m ling. If the wire is subjected to an axial tensile load 10kN what will be its extension?

a) 2.55mm

b) 3.15mm

c) 2.45mm

d) 2.65mm

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Answer: a

Explanation: As change in length = PL/AE

P = 10x 1000N, L = 1m, A = πd2/4 = 1.963 x 10-5 m2, E = 200 x 109 N/m2.

A composite rod is 1000mm long, its two ends are 40mm2 and 30mm2 in area and length are 300mm and 200mm respectively. The middle portion of the rod is 20mm2 in area. If the rod is subjected to an axial tensile load of 1000N, what will be its total elongation (E = 200GPa)?

a) 0.145mm

b) 0.127mm

c) 0.187mm

d) 0.196mm

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Answer: d

Explanation: P = 1000N, Area A1 = 40mm2, A2 = 20mm2, A30 = 30mm2

Length, L1 = 300mm, L2 = 500mm, L3 = 200mm

E = 200GPa = 200x 1000 N/mm2

Total extension = P/E x (L1/A1 + L2/A2 + L3/A3).

A rod of two sections of area 625mm2 and 2500mm2 of length 120cm and 60cm respectively. If the load applied is 45kN then what will be the elongation (E = 2.1x 105 N/mm2)?

a) 0.462mm

b) 0.521mm

c) 0.365mm

d) 0.514mm

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Answer: a

Explanation: P = 45,000N, E =2.1x 105 N/mm2,

Area, A1 = 625mm2, A2 = 2500mm2,

Length, L1 = 1200mm, L2 = 600mm

Elongation = P/E x (L1/A1 + L2/A2).

What will be the elongation of a bar of 1250mm2 area and 90cm length when applied a force of 130kN if E = 1.05x 105 N/mm2?

a) 0.947mm

b) 0.891mm

c) 0.845mm

d) 0.745mm

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Answer: b

Explanation: As change in length = PL/AE

P = 130x 1000N, L = 900mm, A = 1250 mm2, E = 1.05 x 105 N/m2.

A bar shown in diagram is subjected to load 160kN. If the stress in the middle portion Is limited to 150N/mm2, what will be the diameter of the middle portion?<br/>

a) 3.456 cm<br/>

b) 3.685 cm<br/>

c) 4.524 cm<br/>

d) 4.124 cm

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Answer: b

Explanation: Let L2 and D2 be the dimensions of the middle portion and L1 and D2 be the end portion dimensions.

For middle portion area = load / stress

This gives area by which diameter can be calculated.

A steel bar of 20mm x 20mm square cross-section is subjected to an axial compressive load of 100kN. If the length of the bar is 1m and E=200GPa, then what will be the elongation of the bar?

a) 1.25mm

b) 2.70mm

c) 5.40mm

d) 4.05mm

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Answer: a

Explanation: Elongation in bar = PL/ AE = (100x1000x1) / (0.2×0.2x200x106) = 1.25mm.

A solid uniform metal bar is hanging vertically from its upper end. Its elongation will be _________

a) Proportional to L and inversely proportional to D2

b) Proportional to L2 and inversely proportional to D

c) Proportional of U but independent of D

d) Proportional of L but independent of D

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Answer: a

Explanation: Elongation = WL / 2AE = 4WL / 2πD2E α L/D2.

Strength of Materials MCQs Bars of Composite Sections – 2

A member ABCD is subjected to points load P1=45kN, P2, P3=450kN and P4=130kN. what will be the value of P necessary for equilibrium?<br/>

a) 350kN<br/>

b) 365kN<br/>

c) 375kN<br/>

d) 400kN

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Answer: b

Explanation: On resolving forces P1 + P3 = P2 + P4

So P2 = 45 + 450 – 130 I.e. P2 = 365kN.

A member ABCD is subjected to points load P1=45kN, P2, P3=450kN and P4=130kN. What will be the total elongation of the member, assuming the modulus of elasticity to be 2.1x105N/mm2. The cross sectional area is 625mm, 2500mm, 1250mm respectively.

a) 0.4914mm

b) 0.4235mm

c) 0.4621mm

d) 0.4354mm

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Answer: a

Explanation: First of all the fores will be calculated

on resolving forces P1 + P3 = P2 + P4

So P2 = 45 + 450 – 130 I.e. P2 = 365kN

So forces on three sections will be 45kN, 320kN and 130kN respectively.

After that increase in length = PL/AE for all three sections will be calculated.

A tensile rod of 40kN is acting on a rod of diameter 40mm and of length 4m. a bore of diameter 20mm is made centrally on the rod. To what length the rod should be bored so that the total extension will increase 30% under the same tensile load if E = 2×105 N/mm2?<br/>

a) 2m<br/>

b) 2.7m<br/>

c) 3.2m<br/>

d) 3.6m

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Answer: d

Explanation: The extension = PL / AE = 2/π mm

Extension after the bore is made = 1.3x 2/π mm = 2.6/π mm

The extension after the bore is made, is also obtained by finding the extension of the un bored length and bored length.

Stress = load / area

So total extension after bore is made can have two equations which can be put equal and the length the rod should be bored up is calculated.

A bar is subjected to a tensile load of 150kN. If the stress in the middle portion is limited to 160 N/mm2, what will be the diameter of the middle portion of the total elongation of the bar is 0.25cm (E= 2 x 105)?<br/>

a) 3cm<br/>

b) 3.45cm<br/>

c) 3.85cm<br/>

d) 4cm

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Answer: b

Explanation: Total extension = P/E x (L1/A1 + L2/A2 + L3/A3 )

Only variable in the equation is A2. after getting this the diameter of the section can be calculated.

A rod, which tapers uniformly from 5cm diameter to 3cm diameter in a length of 50cm, is subjected to an axial load of 6000N. if E = 2,00,000 N/mm2, what will be the extension of the rod?

a) 0.00114cm

b) 0.00124cm

c) 0.00127cm

d) 0.00154cm

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Answer: c

Explanation: The extension in the rod = PL / Et(a-b) x loge (a/b)

Where a = 50mm, b = 30mm.

A bar is in two sections having equal lengths. The area of cross section of 1st is double that of 2nd. if the bar carries an axial load of P, then what will be the ratio of elongation in section 2nd to section 1st ?

a) 1/2

b) 2

c) 4

d) 1/4

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Answer: b

Explanation: Ratio of elongation in 2nd / ratio of elongation in 1st = L2/L1 x A2/A1

Since L1 = L2 and A1 = 2A2

Therefore, ratio = 1 x 2/1 = 2.

A round bar made of same material consists of 4 parts each of 100mm length having diameters of 40mm, 50mm, 60mm and 70mm, respectively. If the bar is subjected to an axial load of 10kN, what will be the total elongation of the bar in mm?

a) 0.4/πE ( 1/16 + 1/25 + 1/36 + 1/49)

b) 4/πE ( 1/16 + 1/25 + 1/36 + 1/49)

c) 2/πE ( 1/16 + 1/25 + 1/36 + 1/49)

d) 40/πE ( 1/16 + 1/25 + 1/36 + 1/49)

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Answer: d

Explanation: Total elongation = 4PL/πE ( 1/d12 + 1/d22 + 1/d32 + 1/d42)

= 4x10x100/πEx100 ( 1/16 + 1/25 + 1/36 + 1/49) mm

= 40/πE ( 1/16 + 1/25 + 1/36 + 1/49).

A bar shown in the diagram below is subjected to load 160kN. If the stress in the middle portion Is limited to 150N/mm2, what will be the length of the middle portion, if the total elongation of the bar is to be 0.2mm? Take E = 2.1 x 105 N/mm2.<br/>

a) 18.45cm<br/>

b) 17.24cm<br/>

c) 16.45cm<br/>

d) 20.71cm

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Answer: d

Explanation: Let L2 and D2 be the dimensions of the middle portion and L1 and D2 be the end portion dimensions.

For middle portion area = load / stress

This gives area by which diameter can be calculated.

As Total extension = P/E x (L1/A1 + L2/A2)

This gives the value of L2.

A composite bar consists of a bar enclosed inside a tune of another material when compressed under a load as whole through rigid collars at the end of the bar. What will be the equation of compatibility?

a) W1 + W2 = W

b) W1 + W2 = constant

c) W1/A1E1 = W2/A2E2

d) W1/A1E2 = W2/A2E1

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Answer: a

Explanation: Compatibility equation insists that the change in length of the bar must be compatible with the boundary conditions. Here W1 + W2 = W it is also correct but it is equilibrium equation.

### Strength of Materials MCQs Definition of Strain Energy

What is the strain energy stored in a body due to gradually applied load?

a) σE/V

b) σE2/V

c) σV2/E

d) σV2/2E

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Answer: d

Explanation: Strain energy when load is applied gradually = σ2V/2E.

Strain energy stored in a body to uniform stress s of volume V and modulus of elasticity E is __________

a) s2V/2E

b) sV/E

c) sV2/E

d) sV/2E

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Answer: a

Explanation: Strain energy = s2V/2E.

In a material of pure shear stress τthe strain energy stored per unit volume in the elastic, homogeneous isotropic material having elastic constants E and v will be:

a) τ2/E x (1+ v)

b) τ2/E x (1+ v)

c) τ2/2E x (1+ v)

d) τ2/E x (2+ v)

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Answer: a

Explanation: σ1=τ, σ2= -τσ3=0

U = (τ2+- τ2-2μτ(-τ))V = τ2/E x (1+ v)V.

PL3/3EI is the deflection under the load P of a cantilever beam. What will be the strain energy?

a) P2L3/3EI

b) P2L3/6EI

c) P2L3/4EI

d) P2L3/24EI

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Answer: b

Explanation: We may do it taking average

Strain energy = Average force x displacement = (P/2) x PL3/3EI = P2L3/6EI.

A rectangular block of size 400mm x 50mm x 50mm is subjected to a shear stress of 500kg/cm2. If the modulus of rigidity of the material is 1×106 kg/cm2, the strain energy will be __________

a) 125 kg-cm

b) 1000 kg-cm

c) 500 kg-cm

d) 100 kg-cm

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Answer: a

Explanation: Strain energy stored = τ2V/2G = 5002/2×106 x 40x5x5 = 125 kg-cm.

A material of youngs modulus and Poissons ratio of unity is subjected to two principal stresses σ1 and σ2 at a point in two dimensional stress system. The strain energy per unit volume of the material is __________

a) (σ12 + σ22 – 2σ1σ2 ) / 2E

b) (σ12 + σ22 + 2σ1σ2 ) / 2E

c) (σ12 – σ22 – 2σ1σ2 ) / 2E

d) (σ12 – σ22 – 2σ1σ2 ) / 2E

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Answer: a

Explanation: Strain energy = (σ1ε1+ σ1ε1 ) / 2E

= (σ12 + σ22 – 2σ1σ2 ) / 2E.

If forces P, P and P of a system are such that the force polygon does not close, then the system will __________

a) Be in equilibrium

b) Reduce to a resultant force

c) Reduce to a couple

d) Not be in equilibrium

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Answer: d

Explanation: The forces are not concurrent so the resultant force and couple both may be present. Thus the best choice is that forces are not in equilibrium.

The strain energy in a member is proportional to __________

a) Product of stress and the strain

b) Total strain multiplied by the volume of the member

c) The maximum strain multiplied by the length of the member

d) Product of strain and Young’s modulus of the material

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Answer: d

Explanation: Strain energy per unit volume for solid = q2 / 4G.

A bar of cross-section A and length L is subjected to an axial load W. the strain energy stored in the bar would be __________

a) WL / AE

b) W2L / 4AE

c) W2L / 2AE

d) WL / 4AE

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Answer: c

Explanation: Deformation in the bar = WL / AE

Strain energy = W/2 x WL / AE = W2L / 2AE.

A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 5m long. What is the stretch in the rod if E = 2×105 N/mm2?

a) 1.1mm

b) 1.24mm

c) 2mm

d) 1.19mm

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Answer: d

Explanation: Stress = Load/ area = 60,000 / (π/4 D2) = 470746 N/mm2

So stretch = stress x length / E = 1.19mm.

A tensile load of 50kN is gradually applied to a circular bar of 5cm diameter and 5m long. What is the strain energy absorbed by the rod (E = 200GPa)?

a) 14 N-m

b) 15.9 N-mm

c) 15.9 N-m

d) 14 N-mm

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Answer: c

Explanation: Stress = 50,000 / 625π = 25.46

Strain energy = σ2V/2E = 25.46×25.46×9817477 / (2×200000) = 15909.5 N-mm = 15.9 N-m.

A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 5m long. What is the strain energy in the rod if the load is applied suddenly (E = 2×105 N/mm2)?

a) d143.23 N-m

b) 140.51 N-m

c) 135.145 N-m

d) 197.214 N-m

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Answer: a

Explanation: Maximum instantaneous stress = 2P / A = 95.493

Strain energy = σ2V/2E = 143288N-mm = 143.238 N-m.